Does Every Point of R Exist as an Accumulation Point in a Sequence?

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SUMMARY

A sequence can exist where every point of R (real numbers) is an accumulation point, specifically through the set of rational numbers, as they are dense in R. The discussion highlights that the rational numbers are countable, allowing for enumeration. Additionally, the Heine-Borel Theorem is illustrated by demonstrating that a non-closed bounded set, such as the open interval (-1, 1), can be covered by an open cover that lacks a finite sub-cover, emphasizing the necessity of closed sets in this theorem.

PREREQUISITES
  • Understanding of accumulation points in topology
  • Familiarity with the Heine-Borel Theorem
  • Knowledge of countability and enumeration of sets
  • Concept of open covers in real analysis
NEXT STEPS
  • Study the properties of accumulation points in metric spaces
  • Explore the implications of the Heine-Borel Theorem in real analysis
  • Learn about countable and uncountable sets, focusing on rational numbers
  • Investigate open covers and finite sub-covers in topology
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Mathematicians, students of real analysis, and anyone studying topology or the properties of sequences and sets in mathematics.

Kraziethuy
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1.Does a sequence exist that has every point of R(real numbers) as an accumulation point?

2.Show that closed is essential in the Heine-Borel Theorem by finding an open cover of a non-closed bounded set that does not have a finite sub-cover.

I think that the set of rational numbers has every real numbers as an accumulation point, but I'm unsure of the sequence of rational numbers.

Any help appreciated, thanks.
 
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I think that the set of rational numbers has every real numbers as an accumulation point, but I'm unsure of the sequence of rational numbers.

You have the right idea. A potentially unsatisfying answer might be to simply enumerate the rational numbers (which is possible because they are countable).
Alternatively, are you familiar with Hilbert's hotel and the proof that the rational numbers are countable?
 
For Q2, just consider the open interval of (-1, 1) in R. Consider an open cover like the collection of intervals, { (x-w, x+w), w=(1-|x|)/2: x from (-1, 1) }. If this collection has a finite subcover, would there be any points not covered by the subcovering (and thus a contradiction)?
 
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